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Sakurai (Modern Quantum Mechanics, by J.J. Sakurai) states in the section on compatible operators:

Let us first consider the case of compatible observables A and B. As usual, we assume that the ket space is spanned by the eigenkets of A. We may also regard the same ket space as being spanned by the eigenkets of B

Under what conditions can we say that the space spanned by the eigenkets of A can be spanned by that of B? If this is true only for commuting operators (compatible observables), how can we derive the truth of this fact from the definition of compatible observables, i.e. [$A,B] = \hat{0}$?

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    $\begingroup$ Assuming the observables are defined on the same space, why would you need commutativity to say that their eigenkets span the same space? You can just apply the spectral theorem to each separately, no? $\endgroup$ – ACuriousMind Feb 4 '17 at 14:29
  • $\begingroup$ Yes, but are we assuming here that the operators are defined on the same domain space? They should, I guess.. $\endgroup$ – Arkya Chatterjee Feb 4 '17 at 14:31
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This is how I think about it.

There is only one space in question here, the Hilbert space. My system's state depends entirely on the state vector it is in, and this state vector belongs in the Hilbert space.

Now, how do I define this space? I define it by saying that it is the basis formed by the eigenkets of an operator(observable), any operator.

After all, when I try to measure the value of that observable, it causes the system to land in another state vector (which we know should be an eigenket), therefore it makes sense that if two operators are being used on the same system, they should span the same space.

This is therefore not at all related to the compatibility.

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    $\begingroup$ A priori, therefore, we need to know what observable(s), we're interested in? If that is the case, what if there's more internal structure to the space? For example, spin was hypothesized initially because hydrogen spectrum was seen to have fine structures which couldn't be explained within the original Hilbert space. On a similar vein, does our hypothesis of a Hilbert space depend on experiment? This seems slightly unsettling to me. $\endgroup$ – Arkya Chatterjee Dec 10 '17 at 6:04

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